103 Trigonometry Problems

1. Trigonometric Fundamentals 37
The Law of Cosines in Action, Take II: Heron’s Formula
and Brahmagupta’s Formula
[Brahmagupta’s Formula] Let ABCD be a convex cyclic quadrilateral (Figure
1.38). Let |AB| =a, |BC|=b, |CD|=c, |DA| =d, and s = (a + b + c + d)/2.
Then
[ABCD] = √ (s − a)(s − b)(s − c)(s − d).
A
a
d
D
B
c
b
Figure 1.38.
C
Let B = ̸ ABC and D = ̸ ADC. Applying the law of cosines to triangles ABC
and DBC yields
a 2 + b 2 − 2ab cos B = AC 2 = c 2 + d 2 − 2cd cos D.
Because ABCD is cyclic, B + D = 180 ◦ , and so cos B =−cos D. Hence
It follows that
cos B = a2 + b 2 − c 2 − d 2
.
2(ab + cd)
sin 2 B = 1 − cos 2 B = (1 + cos B)(1 − cos B)
=
(1 + a2 + b 2 − c 2 − d 2 )(1 − a2 + b 2 − c 2 − d 2 )
2(ab + cd)
2(ab + cd)
Note that
= a2 + b 2 + 2ab − (c 2 + d 2 − 2cd)
2(ab + cd)
= [(a + b)2 − (c − d) 2 ][(c + d) 2 − (a − b) 2 ]
4(ab + cd) 2 .
· c2 + d 2 + 2cd − (a 2 + b 2 − 2ab)
2(ab + cd)
(a + b) 2 − (c − d) 2 = (a + b + c − d)(a + b + d − c) = 4(s − d)(s − c).